# Integration by parts, quick question

• Jun 17th 2013, 08:18 AM
Brennox
Integration by parts, quick question
So i understand this question mostly, i dont understand where they got the (1+4) from? and shouldnt there be a negative in front of the last part on the 2nd line.

cheers
• Jun 17th 2013, 10:19 AM
ebaines
Re: Integration by parts, quick question
At the end of the second line what you have is:

$\displaystyle \int e^{2t} \cos t dt = -[2e^{2t})(- \cos t)] + \int 4 e^{2t}(- \cos t) dt$

Notice that the last term is common with the left hand side, so move it to the other side to get:

$\displaystyle (1+4) \int e^{2t} \cos t dt = -2e^{2t}( - \cos t)$

And no, the plus sign before the last term of the 2nd line is correct. Starting with

$\displaystyle - \int 2 e ^{2t} \sin t dt$

and using integration by parts, let $\displaystyle u = -2e ^{2t}$, $\displaystyle dv = \sin t dt$,

then:

$\displaystyle \int u dv = uv - \int v du = -2e^{2t}(-\cos t) - \int (-\cos t) (-4) e^{2t} dt$

$\displaystyle = -2e^{2t} (-\cos t) + \int 4 e^{2t} (-\cos t) dt$
• Jun 17th 2013, 10:27 AM
HallsofIvy
Re: Integration by parts, quick question
I agree, that is hard to make sense of- they leave a lot out. The problem is to integrate $\displaystyle \int_{-\pi}^{\pi} e^{2t}cos(t)dt$. They note that "it doesn't matter which function is chosen to differentiate" and then choose to differentiate the $\displaystyle e^{2t}$.

That is we let $\displaystyle u= e^{2t}$ and $\displaystyle dv= cos(t)dt$. Then $\displaystyle du= 2e^{2t}dt$ and $\displaystyle v= sin(t)$.

$\displaystyle uv- \int v du$, then, is $\displaystyle e^{2t}sin(t)\left]_{-\pi}^\pi- 2\int_{-\pi}^\pi e^{2t}sin(t)dt$

Of course, $\displaystyle sin(\pi)= sin(-\pi)=$ so that first term is 0. That's the first "0" in the second line. What's left is $\displaystyle 2\int_{-\pi}^\pi e^{2t}sin(t)dt$.

Now, do the integration by parts again, letting $\displaystyle u= e^{2t}$ and $\displaystyle dv= sin(t)dt$ so that $\displaystyle du= 2e^{2t}dt$ and $\displaystyle v= -cos(t)$.

$\displaystyle 2\int_{-\pi}^\pi e^{2t}sin(t)dt= 2\left( -e^{2t}cos(t)\right]_{-\pi}^\pi- 2\int_{-\pi}^\pi e^{2t}cos(t)dt\right)$

$\displaystyle cos(\pi)= cos(-\pi)= -1$ so the first term is $\displaystyle 2((-1)e^{2\pi}-(-1)e^{-2\pi})= 2(e^{-\pi}- e^{\pi})$
The integral, at this point is $\displaystyle \int e^{2t}cos(t)dt= 2(e^{-\pi}- e^{\pi})- 4\int_{-\pi}^\pi e^{2t}cos(t) dt$

Now, here's the point. Because the first integral takes us from "cos(t)" to "sin(t)" and the second integral takes back to "cos(t)", while the derivatives of $\displaystyle e^{2t}$ always gives "$\displaystyle e^{2t}$", we have come right back (that integral on the right) to what we started with (the integral on the left). So combine those by adding $\displaystyle 4\int_{-\pi}^\pi e^{2t}cos(t)dt$, giving $\displaystyle 5\int_{-\pi}^\pi e^{2t}cox(t)dt$ on the left. That is where the "4+ 1= 5" on the left came from.
• Jun 17th 2013, 03:07 PM
Brennox
Re: Integration by parts, quick question
x1________x2
same______integrate **************(line one)
-(diff______same)

same________integrate
-(diff________same) *******************(line 2)

This is a boss way for noobs like me to remember the order. brackets mean with an integral sign at the front.